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Simple Integral, Still Having Trouble

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    I am trying to take the integral of 1/((-4t2+4t+3)1/2)

    I know that i need to complete the square and it should come out to an inverse sin function but i dont understand how the completed square in the denominator is equal to (4-(2t-1)2)1/2

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 2, 2011 #2
    just expand the square and you see they are equal
  4. Oct 2, 2011 #3
    But how do you come to that? I understand its equal, i just dont know how they calculated it, am I missing something completely obvious or...?
  5. Oct 2, 2011 #4
    have you studied how to complete square ? if you have a quadratic equation

    [tex]ax^2+bx+c = 0[/tex] then you add and subtract middle term square divided by

    4 times the first term




    so use this procedure. in your case there is just a quadratic term, no equation. but
    this approach can still be used
  6. Oct 2, 2011 #5
    Yeah, I have, and for some reason this problem completely stumps me, If you use those formulas , you end up with 4(t+1/2)^2 = -2

    I dont see how that is equivalent to 4-(2t-1)^2
  7. Oct 2, 2011 #6

    [tex]-4t^2+4t +\frac{(4t)^2}{4(-4t^2)}-\frac{(4t)^2}{4(-4t^2)}+3[/tex]





  8. Oct 2, 2011 #7
    Ok, never mind i see that they are equivalent but how do you know to rearrange them in that manner to take the integral? is there any sort of method or is it just through practice
  9. Oct 2, 2011 #8
    oooook..Im sorry ive been so troublesome, thank you so much!
  10. Oct 2, 2011 #9
  11. Oct 2, 2011 #10
    mta, i think you should REALLY know how to complete squares BEFORE you take on calculus. have you had pre-calculus ?
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